was last year's methods exam 1 considered easy?

[S200]

Sorry. See modified working.

[S_R_K]

30 raw is, roughly, 50-55% on each exam (assuming SAC results are solid). Concentrate on getting the marks on questions that are pretty common and have standard techniques, ie. differentiating using product/quotient/chain rule; anti-diff by recognition; finding the rule and domain of a composite function; sketching a graph, including finding stationary points; solving equations involving exponentials or trig ratios, including those that involving making a substitution. You can generally get about 20/40 by doing this, and then give yourself a chance to gain a few marks by writing down something intelligible for tougher questions.

[sailinginwater]

30 raw is, roughly, 50-55% on each exam (assuming SAC results are solid). Concentrate on getting the marks on questions that are pretty common and have standard techniques, ie. differentiating using product/quotient/chain rule; anti-diff by recognition; finding the rule and domain of a composite function; sketching a graph, including finding stationary points; solving equations involving exponentials or trig ratios, including those that involving making a substitution. You can generally get about 20/40 by doing this, and then give yourself a chance to gain a few marks by writing down something intelligible for tougher questions.

Thanks
How about for 35 raw?
And in all honesty, is 1 month enough to get better at certain types of questions I'm struggling with?

[S_R_K]

Can you please explain why part a is p(0) and not (3/5)^1 since if is not successful on the first attempt then he would not try again?

Think about whether your reasoning makes sense. If he doesn't try again after being unsuccessful on the first attempt, then he doesn't really have three attempts. Either he gets in or he doesn't, and that's it, no more tries. So saying that he wouldn't try again after failing on the first attempt doesn't really make sense in the context of the question, where we are told that he has up to three attempts.

Like S200 said – think about what you would do in this situation. If you've forgotten your password, do you stop trying after the first failed attempt? And do you keep trying after you've already logged in?

[S_R_K]

Thanks
How about for 35 raw?
And in all honesty, is 1 month enough to get better at certain types of questions I'm struggling with?

35 raw you probably need at least 70%.

1 month is a decent chunk of time. Remember, you also have other subjects to deal with. If you are struggling with lots of things, don't spread yourself thin. Concentrate on a couple of things that are achievable, get on top of those, and then move on to something else (but don't forget to periodically go back and consolidate things you are already good at; don't let yourself forget what you've gained).

[Orb]

You have two options
1- do nothing, and be certain to not get the mark that you could have gotten

Or,
2- try your best and let the chips fall where they may

Keep grinding bud, worrying never gets you to the finish line

[sailinginwater]

You have two options
1- do nothing, and be certain to not get the mark that you could have gotten

Or,
2- try your best and let the chips fall where they may

Keep grinding bud, worrying never gets you to the finish line

Hey, sorry if I sound rude, but honestly what makes you think im not going to do any work?

[Orb]

You're typing a response to me right now when you could be studying

Jokes aside, the point is - your end study score is affected by a myriad of factors, and one big one is the amount of time you dedicate to working on practice questions, reviewing your mistakes and learning from them.

You make tens of decisions every day, and if you just change half of your decisions from browsing Instagram or Facebook to working on a maths problem, your end study score will thank you for it.

No work is an exaggeration, but when you think about it, every time you procrastinate when you should be studying, you're doing "no work"

[sailinginwater]

Holly shit I just did the 2015 methods exam 1 and got 38/40!!!
Question 10d took me like 10 minutes to do and 5 minutes to figure out how to actually approach it
And got the answer as 6/√3 when theta is pi/3. Is 6/√3 an acceptable form for the answer?
The 2 marks I lost were due to not knowing how to do question 10a and 10b
Does anyone know how to it?
Thanks

[Sine]

-snip-

you usually need to simplify all your answers properly so a radical on the denominator of a fraction should result in loss of marks.

[S200]

you usually need to simplify all your answers properly so a radical on the denominator of a fraction should result in loss of marks.

How would you simplify that then?

That right?

Holly shit I just did the 2015 methods exam 1 and got 38/40!!!
Question 10d took me like 10 minutes to do and 5 minutes to figure out how to actually approach it
And got the answer as 6/√3 when theta is pi/3. Is 6/√3 an acceptable form for the answer?
The 2 marks I lost were due to not knowing how to do question 10a and 10b
Does anyone know how to it?
Thanks

Good Job!

For 10a

You know the unit Circle, right?

So the co-ords would just be (\(2+2\cos{\theta}\), \(2\sin{\theta}\))

For 10b

; The gradient of the line CT, when it crossed the x-axis, should be \(\tan{\theta}\).
The tangent that the question wants is perpendicular to this, and we know that \(m_1 \times m_2=(-1)\).

So, the gradient of the tangent at T should be \(\frac{-1}{\tan{\theta}}\)

Are any of the questions from the 2015 Exam 1 no longer on the study design?

[sailinginwater]

How would you simplify that then?

That right?

That's what the answers said, so yes you're correct 

[S200]

That's what the answers said, so yes you're correct 

Cool. Do you know how to get there?

[sailinginwater]

Cool. Do you know how to get there?

Need to turn CEDB into a trapezium shape, the area of a trapiezium is (a+b)/2)*h

= ((2-2cos(theta)/sin(theta))+(2/sin(theta))/2*(2)

The 2*2 cancels out, and hence it becomes ((2-2cos(theta)/sin(theta))+(2/sin(theta))

Which simplifies to (4-2cos(theta)/sin(theta))

Using the quotient rule to differentiate

Le u = 4-2cos(theta) let v = sin(theta)

Hence the derivative is ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta))/(sin^2(theta))

Derivative = 0 for minimum

Hence ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta))/(sin^2(theta)) = 0

Hence ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta)) = 0

Hence cos(theta)(-4+2cos(theta)) + 2sin^2(theta) = 0

We know that sin^2(theta) + cos^2(theta) = 1

So -4cos(theta) + 2(cos^2(theta)+sin^2(theta)) = 0

So -4cos(theta) + 2 = 0

So cos(theta) = ½

Base = pi/3

Cos is positive in quadrant 1 and 4

And the domain for theta is from 0 to pi/2

Hence theta = pi/3

When be sub pi/3 into (4-2cos(theta)/sin(theta))

We get the minimum area as 2root3

This question took me 10 minutes to do when I was doing it on the exam lol

Hope this helps

[S200]

-Snip-

Need to turn CEDB into a trapezium shape, the area of a trapiezium is (a+b)/2)*h

= ((2-2cos(theta)/sin(theta))+(2/sin(theta))/2*(2)

The 2*2 cancels out, and hence it becomes ((2-2cos(theta)/sin(theta))+(2/sin(theta))

Which simplifies to (4-2cos(theta)/sin(theta))

Using the quotient rule to differentiate

Le u = 4-2cos(theta) let v = sin(theta)

Hence the derivative is ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta))/(sin^2(theta))

Derivative = 0 for minimum

Hence ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta))/(sin^2(theta)) = 0

Hence ((-4cos(theta) + 2cos^2(theta)+2sin^2(theta)) = 0

Hence cos(theta)(-4+2cos(theta)) + 2sin^2(theta) = 0

We know that sin^2(theta) + cos^2(theta) = 1

So -4cos(theta) + 2(cos^2(theta)+sin^2(theta)) = 0

So -4cos(theta) + 2 = 0

So cos(theta) = ½

Base = pi/3

Cos is positive in quadrant 1 and 4

And the domain for theta is from 0 to pi/2

Hence theta = pi/3

When be sub pi/3 into (4-2cos(theta)/sin(theta))

We get the minimum area as 2root3

This question took me 10 minutes to do when I was doing it on the exam lol

Hope this helps

Sweet! I got lost at the bolded step, so thanks!

And yeah, I also forgot that \(\frac {6\sqrt{3}}{3}=2\sqrt{3}\)...